Funkcija (žalia kreivė) ir jos liestinė taške P (mėlyna tiesė). Funkcijos diferencialas yra dy . Diferencialas - funkcijos pokyčio tiesinė pagrindinė dalis. Funkcija y = f(x) , apibrėžta intervale (a, b ), vadinama diferencijuojamąja taške x
∈
{\displaystyle _{\in }}
, jei jos pokytį Δy = f(x + Δx) - f(x) galima išreikšti dviejų dėmenų suma: Δy = AΔx + o(Δx) ; čia A - skaičius, nepriklausantis nuo Δx .
Pavyzdžiui, yra funkcija f(x)=x2 . Tos funkcijos išvestinė yra
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
=
(
x
+
Δ
x
)
2
−
x
2
Δ
x
=
{\displaystyle {\frac {f(x+\Delta x)-f(x)}{\Delta x}}={\frac {(x+\Delta x)^{2}-x^{2}}{\Delta x}}=}
=
x
2
+
2
x
Δ
x
+
(
Δ
x
)
2
−
x
2
Δ
x
=
2
x
Δ
x
+
(
Δ
x
)
2
Δ
x
=
2
x
+
Δ
x
{\displaystyle ={\frac {x^{2}+2x\Delta x+(\Delta x)^{2}-x^{2}}{\Delta x}}={\frac {2x\Delta x+(\Delta x)^{2}}{\Delta x}}=2x+\Delta x}
y
′
=
f
′
(
x
)
=
d
y
d
x
=
lim
Δ
x
→
0
(
2
x
+
Δ
x
)
=
2
x
.
{\displaystyle y'=f'(x)={\frac {dy}{dx}}=\lim _{\Delta x\to 0}(2x+\Delta x)=2x.}
Įstatykime vietoje x kokią nors reikšmę, pavyzdžiui, x=3.
Δy = AΔx + o(Δx) = 2xΔx + (Δx)2 =6Δx + (Δx)2 , čia A = 2x = 6 = f'(x); o(Δx) = (Δx)2 .
Taigi funkcijos pokytis yra Δy = f(x + Δx) - f(x) = AΔx + o(Δx), o diferencialas dy = AΔx = y'Δx = y'dx = f'(x)dx; Δx = dx.
Kitaip tariant
Δ
y
=
2
x
Δ
x
+
(
Δ
x
)
2
,
{\displaystyle \Delta y=2x\Delta x+(\Delta x)^{2},}
d
y
=
2
x
Δ
x
.
{\displaystyle dy=2x\Delta x.}
Pavyzdžiui, jei
x
=
3
,
Δ
x
=
1
,
{\displaystyle x=3,\;\Delta x=1,}
Δ
y
=
2
x
Δ
x
+
(
Δ
x
)
2
=
2
⋅
3
⋅
1
+
1
2
=
7
{\displaystyle \Delta y=2x\Delta x+(\Delta x)^{2}=2\cdot 3\cdot 1+1^{2}=7}
Δ
y
=
(
x
+
Δ
x
)
2
−
x
2
=
(
3
+
1
)
2
−
3
2
=
16
−
9
=
7
,
{\displaystyle \Delta y=(x+\Delta x)^{2}-x^{2}=(3+1)^{2}-3^{2}=16-9=7,}
d
y
=
2
x
Δ
x
=
2
⋅
3
⋅
1
=
6.
{\displaystyle dy=2x\Delta x=2\cdot 3\cdot 1=6.}
Rasime funkcijos
f
(
x
)
=
y
=
x
3
{\displaystyle f(x)=y=x^{3}}
Δ
y
Δ
x
=
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
=
(
x
+
Δ
x
)
3
−
x
3
Δ
x
=
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}={\frac {(x+\Delta x)^{3}-x^{3}}{\Delta x}}=}
=
x
3
+
3
x
2
Δ
x
+
3
x
(
Δ
x
)
2
+
(
Δ
x
)
3
−
x
3
Δ
x
=
3
x
2
Δ
x
+
3
x
(
Δ
x
)
2
+
(
Δ
x
)
3
Δ
x
=
3
x
2
+
3
x
Δ
x
+
(
Δ
x
)
2
.
{\displaystyle ={\frac {x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-x^{3}}{\Delta x}}={\frac {3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}}{\Delta x}}=3x^{2}+3x\Delta x+(\Delta x)^{2}.}
y
′
=
f
′
(
x
)
=
d
y
d
x
=
lim
Δ
x
→
0
(
3
x
2
+
3
x
Δ
x
+
(
Δ
x
)
2
)
=
3
x
2
.
{\displaystyle y'=f'(x)={\frac {dy}{dx}}=\lim _{\Delta x\to 0}(3x^{2}+3x\Delta x+(\Delta x)^{2})=3x^{2}.}
[ keisti ] Tarkime, turime funkciją
S
(
t
)
=
t
2
.
{\displaystyle S(t)=t^{2}.}
S yra nueitas taško kelias, o t yra laikas. Funkcija
S
(
t
)
{\displaystyle S(t)}
t
.
{\displaystyle t.}
S
′
(
t
)
=
d
S
d
t
{\displaystyle S'(t)={\frac {dS}{dt}}}
t . Taigi
v
(
t
)
=
S
′
(
t
)
=
(
t
2
)
′
=
2
t
.
{\displaystyle v(t)=S'(t)=(t^{2})'=2t.}
Pavyzdžiui, jei laikas skaičiuojamas sekundėmis, surasime iš funkcijos
S
(
t
)
=
t
2
{\displaystyle S(t)=t^{2}}
S
(
t
)
=
t
2
=
10
2
=
100
(
m
)
.
{\displaystyle S(t)=t^{2}=10^{2}=100\;(m).}
Surandame taško momentinį greitį po 10 sekundžių:
v
(
t
)
=
2
t
=
2
⋅
10
=
20
(
m
/
s
)
.
{\displaystyle v(t)=2t=2\cdot 10=20\;(m/s).}
Gilesniam suvokimui apie gravitacija galima paskaityti čia https://lt.wikibooks.org/wiki/Gravitacija O štai pavyzdis, kuris patvirtina, kad momentinis greitis paskaičiuotas teisingai. Pagreitis g=1 (m/s)/s. Laikas t=10 s. Rasime atstumą, kurį nukris akmuo per 10 sekundžių. Ir rasime greitį, kurį pasieks akmuo. Oro pasipriešinimas nepaisomas.
h
=
g
⋅
t
2
2
=
1
⋅
10
2
2
=
50
(
m
)
.
{\displaystyle h={\frac {g\cdot t^{2}}{2}}={\frac {1\cdot 10^{2}}{2}}=50\;(m).}
v
=
g
⋅
t
=
1
⋅
10
=
10
(
m
/
s
)
.
{\displaystyle v=g\cdot t=1\cdot 10=10\;(m/s).}
Kaip matome, atstumas pasiektas 2 kartus mažesnis ir greitis 2 kartus mažesnis. Ar tai įrodo, kad formulės teisingos? Tikriausiai taip. Pavyzdžiui, jei g=2 (m/s)/s, tai gauname:
h
=
g
⋅
t
2
2
=
2
⋅
10
2
2
=
100
(
m
)
.
{\displaystyle h={\frac {g\cdot t^{2}}{2}}={\frac {2\cdot 10^{2}}{2}}=100\;(m).}
v
=
g
⋅
t
=
2
⋅
10
=
20
(
m
/
s
)
.
{\displaystyle v=g\cdot t=2\cdot 10=20\;(m/s).}
Ne veltui sakoma, kad antra funkcijos
S
(
t
)
=
t
2
{\displaystyle S(t)=t^{2}}
a
=
g
=
S
″
(
t
)
=
(
t
2
)
″
=
(
2
t
)
′
=
2
(
m
s
2
)
.
{\displaystyle a=g=S''(t)=(t^{2})''=(2t)'=2\;({\frac {m}{s^{2}}}).}
Trumpai, jeigu S (t ) yra atstumas nueitas per laiką t , kai pagreitis yra a , o atstumo priklausomybė nuo laiko užrašoma formule
S
(
t
)
=
a
t
2
2
,
{\displaystyle S(t)={\frac {at^{2}}{2}},}
tai momentinio greičio formulė yra tokia:
S
′
(
t
)
=
(
a
t
2
2
)
′
=
2
a
t
2
=
a
t
=
v
(
t
)
;
{\displaystyle S'(t)=({\frac {at^{2}}{2}})'={\frac {2at}{2}}=at=v(t);}
o pagreitis randamas šitaip:
S
″
(
t
)
=
(
a
t
)
′
=
a
.
{\displaystyle S''(t)=(at)'=a.}
[ keisti ] Sumos diferencijavimas.
[
u
(
x
)
+
v
(
x
)
]
′
=
u
′
(
x
)
+
v
′
(
x
)
{\displaystyle [u(x)+v(x)]'=u'(x)+v'(x)}
[
u
(
x
)
−
v
(
x
)
]
′
=
u
′
(
x
)
−
v
′
(
x
)
{\displaystyle [u(x)-v(x)]'=u'(x)-v'(x)}
Sandaugos diferencijavimas.
[
u
(
x
)
v
(
x
)
]
′
=
u
′
(
x
)
v
(
x
)
+
u
(
x
)
v
′
(
x
)
{\displaystyle [u(x)v(x)]'=u'(x)v(x)+u(x)v'(x)}
Dalmens diferencijavimas.
[
u
(
x
)
v
(
x
)
]
′
=
u
′
(
x
)
v
(
x
)
−
u
(
x
)
v
′
(
x
)
v
2
(
x
)
{\displaystyle [{\frac {u(x)}{v(x)}}]'={\frac {u'(x)v(x)-u(x)v'(x)}{v^{2}(x)}}}
[ keisti ]
f
′
(
g
(
x
)
)
=
f
′
(
t
)
g
′
(
x
)
,
t
=
g
(
x
)
{\displaystyle f'(g(x))=f'(t)g'(x),\,t=g(x)}
f
(
x
)
=
(
x
2
+
1
)
3
{\displaystyle f(x)=(x^{2}+1)^{3}}
kur
f
′
(
t
)
=
(
t
3
)
′
{\displaystyle f'(t)=(t^{3})'}
g
′
(
x
)
=
(
x
2
+
1
)
′
{\displaystyle g'(x)=(x^{2}+1)'}
f
′
(
x
)
=
f
′
(
t
)
g
′
(
x
)
=
(
t
3
)
′
(
x
2
+
1
)
′
=
3
t
2
2
x
=
3
(
x
2
+
1
)
2
2
x
{\displaystyle f'(x)=f'(t)g'(x)=(t^{3})'(x^{2}+1)'=3t^{2}2x=3(x^{2}+1)^{2}2x}
Pavyzdys iš trigonometrijos,
f
(
x
)
=
sin
(
x
2
)
,
{\displaystyle f(x)=\sin(x^{2}),\,}
f
′
(
t
)
=
(
sin
(
t
)
)
′
{\displaystyle f'(t)=(\sin(t))'}
g
′
(
x
)
=
(
x
2
)
′
{\displaystyle g'(x)=(x^{2})'}
f
′
(
x
)
=
f
′
(
t
)
g
′
(
x
)
=
(
sin
(
t
)
)
′
(
x
2
)
′
=
2
x
cos
(
t
)
=
2
x
cos
(
x
2
)
{\displaystyle f'(x)=f'(t)g'(x)=(\sin(t))'(x^{2})'=2x\cos(t)=2x\cos(x^{2})}
Patikriname gautą atsakymą integruodami keičiant kintamajį:
∫
2
x
cos
(
x
2
)
d
x
=
∫
2
x
cos
(
x
2
)
d
(
x
2
)
2
x
=
∫
cos
(
x
2
)
d
(
x
2
)
=
sin
(
x
2
)
,
{\displaystyle \int 2x\cos(x^{2})\;dx=\int 2x\cos(x^{2})\;{\frac {d(x^{2})}{2x}}=\int \cos(x^{2})\;d(x^{2})=\sin(x^{2}),}
čia
d
(
x
2
)
=
2
x
d
x
,
{\displaystyle d(x^{2})=2x\;dx,}
d
x
=
d
(
x
2
)
2
x
.
{\displaystyle dx={\frac {d(x^{2})}{2x}}.}
[ keisti ]
f
(
x
)
=
g
(
x
)
h
(
x
)
=
e
ln
g
(
x
)
h
(
x
)
=
e
h
(
x
)
ln
g
(
x
)
,
{\displaystyle f(x)=g(x)^{h(x)}=e^{\ln g(x)^{h(x)}}=e^{h(x)\ln g(x)},}
f
′
(
x
)
=
(
g
(
x
)
h
(
x
)
)
′
=
(
e
h
(
x
)
ln
g
(
x
)
)
′
=
e
h
(
x
)
ln
g
(
x
)
(
h
(
x
)
ln
g
(
x
)
)
′
=
g
(
x
)
h
(
x
)
⋅
(
h
(
x
)
ln
g
(
x
)
)
′
.
{\displaystyle f'(x)=(g(x)^{h(x)})'=(e^{h(x)\ln g(x)})'=e^{h(x)\ln g(x)}(h(x)\ln g(x))'=g(x)^{h(x)}\cdot (h(x)\ln g(x))'.}
(
g
(
x
)
h
(
x
)
)
′
=
g
(
x
)
h
(
x
)
⋅
(
h
′
(
x
)
ln
g
(
x
)
+
h
(
x
)
g
′
(
x
)
g
(
x
)
)
.
{\displaystyle (g(x)^{h(x)})'=g(x)^{h(x)}\cdot (h'(x)\ln g(x)+h(x){\frac {g'(x)}{g(x)}}).}
Pavyzdžiai
(
(
3
x
4
)
5
x
3
)
′
=
(
(
3
x
4
)
5
x
3
)
(
15
x
2
ln
(
3
x
4
)
+
5
x
3
⋅
12
x
3
3
x
4
)
.
{\displaystyle ((3x^{4})^{5x^{3}})'=((3x^{4})^{5x^{3}})(15x^{2}\ln(3x^{4})+5x^{3}\cdot {\frac {12x^{3}}{3x^{4}}}).}
(
(
3
x
2
+
2
x
+
5
)
8
x
3
+
2
x
2
+
4
)
′
=
{\displaystyle ((3x^{2}+2x+5)^{8x^{3}+2x^{2}+4})'=}
=
(
3
x
2
+
2
x
+
5
)
8
x
3
+
2
x
2
+
4
(
(
24
x
2
+
4
x
)
ln
(
3
x
2
+
2
x
+
5
)
+
(
8
x
3
+
2
x
2
+
4
)
6
x
+
2
3
x
2
+
2
x
+
5
)
.
{\displaystyle =(3x^{2}+2x+5)^{8x^{3}+2x^{2}+4}((24x^{2}+4x)\ln(3x^{2}+2x+5)+(8x^{3}+2x^{2}+4){\frac {6x+2}{3x^{2}+2x+5}}).}
(
(
sin
x
)
cos
x
)
′
=
(
sin
x
)
cos
x
(
cos
2
x
sin
x
−
sin
x
ln
sin
x
)
.
{\displaystyle ((\sin x)^{\cos x})'=(\sin x)^{\cos x}({\frac {\cos ^{2}x}{\sin x}}-\sin x\ln \sin x).}
(
3
x
+
5
4
x
+
7
)
′
=
3
x
+
5
4
x
+
7
⋅
(
1
2
(
ln
(
3
x
+
5
)
−
ln
(
4
x
+
7
)
)
)
′
=
3
x
+
5
4
x
+
7
⋅
1
2
(
3
3
x
+
5
−
4
4
x
+
7
)
.
{\displaystyle ({\sqrt {\frac {3x+5}{4x+7}}})'={\sqrt {\frac {3x+5}{4x+7}}}\cdot ({\frac {1}{2}}(\ln(3x+5)-\ln(4x+7)))'={\sqrt {\frac {3x+5}{4x+7}}}\cdot {\frac {1}{2}}({\frac {3}{3x+5}}-{\frac {4}{4x+7}}).}
(
x
1
x
)
′
=
x
1
x
⋅
(
1
x
ln
x
)
′
=
x
1
x
⋅
(
−
1
x
2
ln
x
+
1
x
⋅
1
x
)
=
x
1
x
⋅
1
−
ln
x
x
2
.
{\displaystyle (x^{\frac {1}{x}})'=x^{\frac {1}{x}}\cdot ({\frac {1}{x}}\ln x)'=x^{\frac {1}{x}}\cdot (-{\frac {1}{x^{2}}}\ln x+{\frac {1}{x}}\cdot {\frac {1}{x}})=x^{\frac {1}{x}}\cdot {\frac {1-\ln x}{x^{2}}}.}
[ keisti ] Iš diferencialo apibrėžimo seka, kad jis priklauso linijiniai nuo
Δ
x
{\displaystyle \Delta x}
Δ
y
{\displaystyle \Delta y}
Δ
y
{\displaystyle \Delta y}
Δ
x
{\displaystyle \Delta x}
f
(
x
)
=
x
3
,
{\displaystyle f(x)=x^{3},}
Δ
y
=
(
x
0
+
Δ
x
)
3
−
x
0
3
=
3
x
0
2
Δ
x
+
3
x
0
(
Δ
x
)
2
+
(
Δ
x
)
3
,
{\displaystyle \Delta y=(x_{0}+\Delta x)^{3}-x_{0}^{3}=3x_{0}^{2}\Delta x+3x_{0}(\Delta x)^{2}+(\Delta x)^{3},}
d
y
=
f
′
(
x
0
)
Δ
x
=
(
lim
Δ
x
→
0
(
x
0
+
Δ
x
)
3
−
x
0
3
Δ
x
)
Δ
x
=
3
x
0
2
Δ
x
.
{\displaystyle dy=f'(x_{0})\Delta x={\Big (}\lim _{\Delta x\to 0}{\frac {(x_{0}+\Delta x)^{3}-x_{0}^{3}}{\Delta x}}{\Big )}\Delta x=3x_{0}^{2}\Delta x.}
Daugelyje uždavinių funkcijos priaugimą duotajame taške apytiksliai pakeičia diferencialu šitame taške:
Δ
y
≈
d
y
=
f
′
(
x
0
)
Δ
x
.
{\displaystyle \Delta y\approx dy=f'(x_{0})\Delta x.}
Absoliuti paklaida, padarius tokį pakeitimą, lygi
|
Δ
y
−
d
y
|
{\displaystyle |\Delta y-dy|}
Δ
x
→
0
,
{\displaystyle \Delta x\to 0,}
Δ
x
.
{\displaystyle \Delta x.}
Pavyzdys . Parodysime, kad jeigu
α
{\displaystyle \alpha }
1
+
α
≈
1
+
α
/
2.
{\displaystyle {\sqrt {1+\alpha }}\approx 1+\alpha /2.}
Sprendimas . Nagrinėkime funkciją
f
(
x
)
=
x
.
{\displaystyle f(x)={\sqrt {x}}.}
Δ
x
{\displaystyle \Delta x}
Δ
y
=
x
0
+
Δ
x
−
x
0
≈
d
y
,
{\displaystyle \Delta y={\sqrt {x_{0}+\Delta x}}-{\sqrt {x_{0}}}\approx dy,}
x
0
+
Δ
x
−
x
0
≈
(
x
)
′
|
x
=
x
0
Δ
x
=
(
lim
Δ
x
→
0
x
0
+
Δ
x
−
x
0
Δ
x
)
Δ
x
=
{\displaystyle {\sqrt {x_{0}+\Delta x}}-{\sqrt {x_{0}}}\approx ({\sqrt {x}})'|_{x=x_{0}}\Delta x={\big (}\lim _{\Delta x\to 0}{\frac {{\sqrt {x_{0}+\Delta x}}-{\sqrt {x_{0}}}}{\Delta x}}{\big )}\Delta x=}
=
(
lim
Δ
x
→
0
(
x
0
+
Δ
x
−
x
0
)
(
x
0
+
Δ
x
+
x
0
)
Δ
x
(
x
0
+
Δ
x
+
x
0
)
)
Δ
x
=
(
lim
Δ
x
→
0
(
x
0
+
Δ
x
)
−
x
0
Δ
x
(
x
0
+
Δ
x
+
x
0
)
)
Δ
x
=
{\displaystyle ={\big (}\lim _{\Delta x\to 0}{\frac {({\sqrt {x_{0}+\Delta x}}-{\sqrt {x_{0}}})({\sqrt {x_{0}+\Delta x}}+{\sqrt {x_{0}}})}{\Delta x({\sqrt {x_{0}+\Delta x}}+{\sqrt {x_{0}}})}}{\big )}\Delta x={\Big (}\lim _{\Delta x\to 0}{\frac {(x_{0}+\Delta x)-x_{0}}{\Delta x({\sqrt {x_{0}+\Delta x}}+{\sqrt {x_{0}}})}}{\Big )}\Delta x=}
=
(
lim
Δ
x
→
0
Δ
x
Δ
x
(
x
0
+
Δ
x
+
x
0
)
)
Δ
x
=
(
lim
Δ
x
→
0
1
x
0
+
Δ
x
+
x
0
)
Δ
x
=
1
2
x
0
Δ
x
.
{\displaystyle ={\Big (}\lim _{\Delta x\to 0}{\frac {\Delta x}{\Delta x({\sqrt {x_{0}+\Delta x}}+{\sqrt {x_{0}}})}}{\Big )}\Delta x={\Big (}\lim _{\Delta x\to 0}{\frac {1}{{\sqrt {x_{0}+\Delta x}}+{\sqrt {x_{0}}}}}{\Big )}\Delta x={\frac {1}{2{\sqrt {x_{0}}}}}\Delta x.}
Taigi,
x
0
+
Δ
x
−
x
0
≈
1
2
x
0
Δ
x
.
{\displaystyle {\sqrt {x_{0}+\Delta x}}-{\sqrt {x_{0}}}\approx {\frac {1}{2{\sqrt {x_{0}}}}}\Delta x.}
Iš kur parinkę
x
0
=
1
,
{\displaystyle x_{0}=1,\;}
Δ
x
=
α
,
{\displaystyle \Delta x=\alpha ,}
1
+
α
−
1
≈
1
2
1
α
,
{\displaystyle {\sqrt {1+\alpha }}-{\sqrt {1}}\approx {\frac {1}{2{\sqrt {1}}}}\alpha ,}
1
+
α
−
1
≈
1
2
α
,
{\displaystyle {\sqrt {1+\alpha }}-1\approx {\frac {1}{2}}\alpha ,}
1
+
α
≈
1
+
α
2
.
{\displaystyle {\sqrt {1+\alpha }}\approx 1+{\frac {\alpha }{2}}.}
Pavyzdžiui, jei
α
=
0.0003
,
{\displaystyle \alpha =0.0003,}
1.0003
≈
1
+
0.0003
2
=
1.00015.
{\displaystyle {\sqrt {1.0003}}\approx 1+{\frac {0.0003}{2}}=1.00015.}
1.0003
≈
1.00014998875.
{\displaystyle {\sqrt {1.0003}}\approx 1.00014998875.}
Pavyzdžiui, jei
α
=
0.3
,
{\displaystyle \alpha =0.3,}
1.3
≈
1
+
0.3
2
=
1.15.
{\displaystyle {\sqrt {1.3}}\approx 1+{\frac {0.3}{2}}=1.15.}
1.3
≈
1.140175425.
{\displaystyle {\sqrt {1.3}}\approx 1.140175425.}
[ keisti ]
y
′
=
y
x
,
{\displaystyle y'={y \over x},}
d
y
d
x
=
y
x
,
{\displaystyle {dy \over dx}={y \over x},}
d
y
d
x
=
y
x
,
{\displaystyle {dy \over dx}={y \over x},}
d
y
y
=
d
x
x
,
{\displaystyle {dy \over y}={dx \over x},}
∫
d
y
y
=
∫
d
x
x
,
{\displaystyle \int {dy \over y}=\int {dx \over x},}
ln
|
y
|
=
ln
|
x
|
+
|
C
1
|
,
{\displaystyle \ln |y|=\ln |x|+|C_{1}|,}
|
y
|
=
|
C
1
|
|
x
|
,
{\displaystyle |y|=|C_{1}||x|,}
y
=
±
C
1
x
,
±
C
1
=
C
,
{\displaystyle y=\pm C_{1}x,\;\pm C_{1}=C,}
y
=
C
x
.
{\displaystyle y=Cx.}
y
′
=
−
y
x
,
{\displaystyle y'=-{y \over x},}
d
y
d
x
=
−
y
x
,
{\displaystyle {dy \over dx}=-{y \over x},}
d
y
y
=
−
d
x
x
,
{\displaystyle {dy \over y}=-{dx \over x},}
∫
d
y
y
=
−
∫
d
x
x
,
{\displaystyle \int {dy \over y}=-\int {dx \over x},}
ln
|
y
|
=
−
ln
|
x
|
+
ln
|
C
1
|
=
ln
|
C
1
x
|
,
{\displaystyle \ln |y|=-\ln |x|+\ln |C_{1}|=\ln |{C_{1} \over x}|,}
|
y
|
=
|
C
1
|
|
x
|
,
{\displaystyle |y|={|C_{1}| \over |x|},}
y
=
±
C
1
x
=
C
x
.
{\displaystyle y=\pm {C_{1} \over x}={C \over x}.}
y
′
=
3
y
2
3
,
{\displaystyle y'=3y^{2 \over 3},}
y
′
3
y
2
3
=
1
,
{\displaystyle {y' \over 3y^{2 \over 3}}=1,}
d
y
d
x
3
y
2
3
=
1
,
{\displaystyle {{dy \over dx} \over 3y^{2 \over 3}}=1,}
d
y
3
y
2
3
=
d
x
,
{\displaystyle {dy \over 3y^{2 \over 3}}=dx,}
∫
d
y
3
y
2
3
=
∫
d
x
,
{\displaystyle \int {dy \over 3y^{2 \over 3}}=\int dx,}
1
3
∫
y
−
2
3
d
y
=
∫
d
x
,
{\displaystyle {1 \over 3}\int y^{-{2 \over 3}}dy=\int dx,}
1
3
⋅
y
1
3
1
3
=
x
+
C
,
{\displaystyle {1 \over 3}\cdot {y^{1 \over 3} \over {1 \over 3}}=x+C,}
y
1
3
=
x
+
C
,
{\displaystyle y^{1 \over 3}=x+C,}
y
=
(
x
+
C
)
3
.
{\displaystyle y=(x+C)^{3}.}
y
′
−
2
x
y
2
+
5
=
0
,
{\displaystyle y'-2xy^{2}+5=0,}
d
y
d
x
=
2
x
y
2
−
5
,
{\displaystyle {dy \over dx}=2xy^{2}-5,}
∫
d
y
=
∫
(
2
x
y
2
−
5
)
d
x
,
{\displaystyle \int dy=\int (2xy^{2}-5)dx,}
y
=
x
2
y
2
−
5
x
+
C
.
{\displaystyle y=x^{2}y^{2}-5x+C.}
(
1
+
x
)
y
d
x
+
(
1
−
y
)
x
d
y
=
0
,
{\displaystyle (1+x)ydx+(1-y)xdy=0,}
1
+
x
x
d
x
+
1
−
y
y
=
0
,
{\displaystyle {1+x \over x}dx+{1-y \over y}=0,}
∫
(
1
x
+
1
)
d
x
+
∫
(
1
y
−
1
)
d
y
=
∫
0
d
x
,
{\displaystyle \int ({1 \over x}+1)dx+\int ({1 \over y}-1)dy=\int 0dx,}
ln
|
x
|
+
x
+
ln
|
y
|
−
y
=
C
,
{\displaystyle \ln |x|+x+\ln |y|-y=C,}
ln
|
x
y
|
+
x
−
y
=
C
.
{\displaystyle \ln |xy|+x-y=C.}
y
′
x
−
x
2
−
y
=
0
,
{\displaystyle y'x-x^{2}-y=0,}
d
y
d
x
=
x
+
y
x
,
{\displaystyle {dy \over dx}=x+{y \over x},}
∫
d
y
=
∫
(
x
+
y
x
)
d
x
,
{\displaystyle \int dy=\int (x+{y \over x})dx,}
y
=
x
2
2
+
y
ln
|
x
|
+
C
;
{\displaystyle y={x^{2} \over 2}+y\ln |x|+C;}
y
=
x
2
+
C
x
.
{\displaystyle y=x^{2}+Cx.}
(
x
2
y
+
y
)
d
x
+
(
x
2
−
x
2
y
)
d
y
=
0
,
{\displaystyle (x^{2}y+y)dx+(x^{2}-x^{2}y)dy=0,}
y
(
x
2
+
1
)
d
x
=
x
2
(
y
−
1
)
d
y
,
{\displaystyle y(x^{2}+1)dx=x^{2}(y-1)dy,}
1
+
x
2
x
2
d
x
=
y
−
1
y
d
y
,
{\displaystyle {1+x^{2} \over x^{2}}dx={y-1 \over y}dy,}
∫
(
1
x
2
+
1
)
d
x
=
∫
(
1
−
1
y
)
d
y
+
C
,
{\displaystyle \int ({1 \over x^{2}}+1)dx=\int (1-{1 \over y})dy+C,}
x
−
1
x
−
y
+
ln
|
y
|
=
C
.
{\displaystyle x-{1 \over x}-y+\ln |y|=C.}
y
′
=
2
y
,
{\displaystyle y'=2{\sqrt {y}},}
d
y
d
x
=
2
y
,
{\displaystyle {dy \over dx}=2{\sqrt {y}},}
d
y
2
y
=
d
x
,
{\displaystyle {dy \over 2{\sqrt {y}}}=dx,}
1
2
∫
d
y
y
=
∫
d
x
,
{\displaystyle {1 \over 2}\int {dy \over {\sqrt {y}}}=\int dx,}
1
2
⋅
y
1
2
=
x
+
C
,
{\displaystyle {1 \over 2}\cdot {{\sqrt {y}} \over {1 \over 2}}=x+C,}
y
=
x
+
C
,
{\displaystyle {\sqrt {y}}=x+C,}
y
=
(
x
+
C
)
2
.
{\displaystyle y=(x+C)^{2}.}
y
′
=
x
(
y
2
+
1
)
,
{\displaystyle y'=x(y^{2}+1),}
d
y
y
2
+
1
=
x
d
x
,
y
2
+
1
≠
0
,
{\displaystyle {dy \over y^{2}+1}=xdx,\;y^{2}+1\neq 0,}
arctan
y
=
x
2
2
+
C
,
{\displaystyle \arctan y={x^{2} \over 2}+C,}
y
=
tan
(
x
2
2
+
C
)
,
−
π
2
<
x
2
2
+
C
<
π
2
.
{\displaystyle y=\tan({x^{2} \over 2}+C),\;-{\pi \over 2}<{x^{2} \over 2}+C<{\pi \over 2}.}
y
′
=
1
+
y
2
1
+
x
2
,
{\displaystyle y'={1+y^{2} \over 1+x^{2}},}
y
(
1
)
=
0
,
{\displaystyle y(1)=0,}
d
y
1
+
y
2
=
d
x
1
+
x
2
,
{\displaystyle {dy \over 1+y^{2}}={dx \over 1+x^{2}},}
arctan
y
=
arctan
x
+
C
,
{\displaystyle \arctan y=\arctan x+C,}
arctan
0
=
arctan
1
+
C
,
{\displaystyle \arctan 0=\arctan 1+C,}
arctan
y
=
arctan
x
−
π
4
,
{\displaystyle \arctan y=\arctan x-{\pi \over 4},}
arctan
y
=
arctan
x
−
arctan
1
,
{\displaystyle \arctan y=\arctan x-\arctan 1,}
arctan
y
=
arctan
x
−
1
1
+
x
⋅
1
,
{\displaystyle \arctan y=\arctan {x-1 \over 1+x\cdot 1},}
y
=
x
−
1
x
+
1
.
{\displaystyle y={x-1 \over x+1}.}
Aiškiau, tai reikia pasinaudoti šita formule
tan
(
x
−
y
)
=
tan
x
−
tan
y
1
+
tan
x
tan
y
{\displaystyle \tan(x-y)={\frac {\tan x-\tan y}{1+\tan x\tan y}}}
arctan
y
=
arctan
x
−
π
4
,
{\displaystyle \arctan y=\arctan x-{\pi \over 4},}
tan
arctan
y
=
tan
(
arctan
x
−
π
4
)
=
tan
(
arctan
x
)
−
tan
π
4
1
+
tan
(
arctan
x
)
⋅
tan
π
4
=
x
−
1
1
+
x
⋅
1
,
{\displaystyle \tan \arctan y=\tan(\arctan x-{\pi \over 4})={\frac {\tan(\arctan x)-\tan {\pi \over 4}}{1+\tan(\arctan x)\cdot \tan {\pi \over 4}}}={\frac {x-1}{1+x\cdot 1}},}
y
=
x
−
1
x
+
1
.
{\displaystyle y={x-1 \over x+1}.}
x
y
′
+
y
=
0
,
{\displaystyle xy'+y=0,}
y
|
x
=
3
=
2
;
{\displaystyle y|_{x=3}=2;}
x
d
y
d
x
=
−
y
;
{\displaystyle x{dy \over dx}=-y;}
d
y
y
=
−
d
x
x
;
{\displaystyle {dy \over y}=-{dx \over x};}
ln
|
y
|
=
−
ln
|
x
|
+
ln
|
C
|
;
{\displaystyle \ln |y|=-\ln |x|+\ln |C|;}
y
=
C
x
;
{\displaystyle y={C \over x};}
2
=
C
3
;
{\displaystyle 2={C \over 3};}
C
=
6
;
{\displaystyle C=6;}
y
=
6
x
.
{\displaystyle y={6 \over x}.}
y
′
−
2
x
y
1
+
x
2
=
0
,
{\displaystyle y'-{2xy \over 1+x^{2}}=0,}
d
y
d
x
=
2
x
y
1
+
x
2
,
{\displaystyle {dy \over dx}={2xy \over 1+x^{2}},}
∫
d
y
y
=
∫
2
x
1
+
x
2
d
x
,
{\displaystyle \int {dy \over y}=\int {2x \over 1+x^{2}}dx,}
∫
d
y
y
=
∫
d
(
1
+
x
2
)
1
+
x
2
,
{\displaystyle \int {dy \over y}=\int {d(1+x^{2}) \over 1+x^{2}},}
ln
|
y
|
=
ln
|
1
+
x
2
|
+
ln
|
C
|
,
{\displaystyle \ln |y|=\ln |1+x^{2}|+\ln |C|,}
y
=
C
(
1
+
x
2
)
.
{\displaystyle y=C(1+x^{2}).}
[ keisti ] Uždavinys apie radiaktyvųjį skilimą. Bandymais nustatyta, kad radioaktyviosios medžiagos skilimo greitis proporcingas nesuskilusios medžiagos kiekiui. Nustatykime nesuskilusios medžiagos kiekio m priklausomybę nuo laiko t , kai pradinis medžiagos kiekis lygus
m
0
.
{\displaystyle m_{0}.}
Sprendimas. Radioaktyvios medžiagos skilimo greitis lygus
d
m
d
t
.
{\displaystyle {dm \over dt}.}
Pagal sąlyga,
d
m
d
t
=
−
k
m
;
(
1
)
{\displaystyle {dm \over dt}=-km;\;(1)}
čia
k
>
0
{\displaystyle k>0}
d
m
d
t
<
0
,
{\displaystyle {dm \over dt}<0,}
d
m
m
=
−
k
d
t
,
{\displaystyle {dm \over m}=-k\;dt,}
∫
d
m
m
=
−
k
∫
d
t
,
{\displaystyle \int {dm \over m}=-k\int dt,}
ln
m
=
−
k
t
+
ln
C
.
(
2
)
{\displaystyle \ln m=-kt+\ln C.\;(2)}
(Norėdami supaprastinti bendrojo sprendimo išraišką, vietoje C rašome
ln
C
{\displaystyle \ln C}
ln
m
−
ln
C
=
ln
m
C
=
−
k
t
,
{\displaystyle \ln m-\ln C=\ln {m \over C}=-kt,}
e
ln
m
C
=
m
C
=
e
−
k
t
,
{\displaystyle e^{\ln {m \over C}}={m \over C}=e^{-kt},}
m
=
C
e
−
k
t
.
(
3
)
{\displaystyle m=Ce^{-kt}.\;(3)}
Uždavinio sąlygoje nurodyta, kad laiko momentu
t
=
0
{\displaystyle t=0}
m
0
.
{\displaystyle m_{0}.}
m
|
t
=
0
=
m
0
.
{\displaystyle m|_{t=0}=m_{0}.}
Į (3) reiškinį įrašę vietoje m dydį
m
0
{\displaystyle m_{0}}
t - nulį, gauname:
m
0
=
C
e
−
k
⋅
0
=
C
.
{\displaystyle m_{0}=Ce^{-k\cdot 0}=C.}
m
=
m
0
e
−
k
t
,
{\displaystyle m=m_{0}e^{-kt},}
atitinkantis duotąsias pradines sąlygas.
Radiaktyvios medžiagos skilimo greitį apibūdina pusamžio, arba pusėjimo trukmės, sąvoka. Taip vadinamas laikas T , per kurį suskyla pusė pradinio radiaktyvios medžiagos kiekio. Į (4) lygybę įrašę
m
=
m
0
2
{\displaystyle m={m_{0} \over 2}}
t
=
T
,
{\displaystyle t=T,}
1
2
m
0
=
m
0
e
−
k
T
,
{\displaystyle {1 \over 2}m_{0}=m_{0}e^{-kT},}
1
2
=
e
−
k
T
,
{\displaystyle {1 \over 2}=e^{-kT},}
ln
1
2
=
−
k
T
,
{\displaystyle \ln {1 \over 2}=-kT,}
ln
1
−
ln
2
=
−
k
T
,
{\displaystyle \ln 1-\ln 2=-kT,}
ln
2
−
ln
1
=
ln
2
1
=
ln
2
=
k
T
.
{\displaystyle \ln 2-\ln 1=\ln {2 \over 1}=\ln 2=kT.}
Taigi koeficientą k su skilimo pusamžiu T sieja sąlyga
k
=
ln
2
T
.
{\displaystyle k={\ln 2 \over T}.}
Tuomet atskirasis diferencialinės lygties sprendinys, iš lygties (4), užrašomas taip:
m
=
m
0
e
−
t
ln
2
T
.
{\displaystyle m=m_{0}e^{-t{\ln 2 \over T}}.}
Puamžio T reikšmės nustatomos eksperimentais. Įvairių radioaktyviųjų medžiagų skilimo pusamžio reikšmės yra labai skirtingos (žr. lentelę).
Radiaktyvioji medžiaga
Anglis - 14
Plutonis - 244
Radis - 226
Stroncis - 90
Uranas - 235
Uranas - 238
Skilimo pusamžis
5730 metų
8
⋅
10
7
{\displaystyle 8\cdot 10^{7}}
1575 metų
28.1 metų
7.07
⋅
10
8
{\displaystyle 7.07\cdot 10^{8}}
4.51
⋅
10
9
{\displaystyle 4.51\cdot 10^{9}}
Iš lygties (4), koeficientas k nustatomas iš stebėjimų. Tegu per laiką
t
0
{\displaystyle t_{0}}
a % radiaktyviosios medžiagos nuo pradinės masės
m
0
.
{\displaystyle m_{0}.}
(
1
−
a
100
)
m
0
=
m
0
e
−
k
t
0
,
{\displaystyle (1-{a \over 100})m_{0}=m_{0}e^{-kt_{0}},}
(
1
−
a
100
)
=
e
−
k
t
0
,
{\displaystyle (1-{a \over 100})=e^{-kt_{0}},}
ln
(
1
−
a
100
)
=
−
k
t
0
,
{\displaystyle \ln(1-{a \over 100})=-kt_{0},}
k
=
−
1
t
0
ln
(
1
−
a
100
)
.
{\displaystyle k=-{1 \over t_{0}}\ln(1-{a \over 100}).}
Tokiu budu buvo nustatyta, kad Radžiui
k
=
0.00044
{\displaystyle k=0.00044}
Įstatę šią reikšmę į formulę (4), gauname:
m
=
m
0
e
−
0.00044
t
.
{\displaystyle m=m_{0}e^{-0.00044t}.}
Rasime periodą pusės suskilimo radžio, t. y. laiko tarpą, per kurį suskyla pusė pirminės masės radžio. Įstatę (4) formulę vietoje m reikšmę
m
0
2
,
{\displaystyle {m_{0} \over 2},}
T nustatymui:
m
0
2
=
m
0
e
−
0.00044
T
,
{\displaystyle {m_{0} \over 2}=m_{0}e^{-0.00044T},}
−
ln
2
=
−
0.00044
T
,
{\displaystyle -\ln 2=-0.00044T,}
T
=
ln
2
0.00044
=
0.69314718
0.00044
≈
1575
{\displaystyle T={\ln 2 \over 0.00044}={0.69314718 \over 0.00044}\approx 1575}
Pavyzdžiui, viename sename vadovėlyje radžio skilimo pusamžis yra
T
≈
1550
{\displaystyle T\approx 1550}
k
=
ln
2
1550
≈
0.000447.
{\displaystyle k={\frac {\ln 2}{1550}}\approx 0.000447.}
Iš pradinės radžio masės
m
0
{\displaystyle m_{0}}
t =1000000) liks tik
m
(
t
)
=
m
(
10
6
)
=
m
0
e
−
k
t
=
m
0
e
−
0.000447
⋅
10
6
=
m
0
e
−
447
≈
0.6
⋅
10
−
194
m
0
.
{\displaystyle m(t)=m(10^{6})=m_{0}e^{-kt}=m_{0}e^{-0.000447\cdot 10^{6}}=m_{0}e^{-447}\approx 0.6\cdot 10^{-194}m_{0}.}
Tą patį rezultatą galima gauti
1
2
{\displaystyle {\frac {1}{2}}}
p
=
10
6
1550
=
645.16129
{\displaystyle p={\frac {10^{6}}{1550}}=645.16129\;}
Taigi,
m
p
o
.
m
i
l
i
j
.
=
m
0
⋅
(
1
2
)
p
=
m
0
⋅
(
1
2
)
645.16129
=
6.124908027
⋅
10
−
195
m
0
.
{\displaystyle m_{po.milij.}=m_{0}\cdot ({\frac {1}{2}})^{p}=m_{0}\cdot ({\frac {1}{2}})^{645.16129}=6.124908027\cdot 10^{-195}m_{0}.}
Tiksliai skaičiuojant su
e
=
2.7182818284
{\displaystyle e=2.7182818284}
m
0
e
−
447
=
m
0
⋅
2.7182818284
−
447
=
7.419362548
⋅
10
−
195
m
0
.
{\displaystyle m_{0}e^{-447}=m_{0}\cdot 2.7182818284^{-447}=7.419362548\cdot 10^{-195}m_{0}.}
Dar tiksliau skaičiuojant, reikia tikslesnio
k
=
ln
2
1550
=
0.6931471805599
1550
=
0.00044719172939.
{\displaystyle k={\frac {\ln 2}{1550}}={\frac {0.6931471805599}{1550}}=0.00044719172939.}
m
0
e
−
447.19172939
=
m
0
⋅
2.718281828459
−
447.19172939
=
6.1249080486
⋅
10
−
195
m
0
.
{\displaystyle m_{0}e^{-447.19172939}=m_{0}\cdot 2.718281828459^{-447.19172939}=6.1249080486\cdot 10^{-195}m_{0}.}
Dar protingiau mąstant, galima pastebėti, kad
e
ln
2
=
2.
{\displaystyle e^{\ln 2}=2.}
m
(
t
)
=
m
(
10
6
)
=
m
0
e
−
k
t
=
m
0
e
−
ln
2
1550
⋅
10
6
=
m
0
(
e
ln
2
)
−
10
6
1550
=
m
0
2
−
10
6
1550
=
m
0
2
−
t
T
=
m
0
2
−
p
.
{\displaystyle m(t)=m(10^{6})=m_{0}e^{-kt}=m_{0}e^{-{\frac {\ln 2}{1550}}\cdot 10^{6}}=m_{0}(e^{\ln 2})^{-{\frac {10^{6}}{1550}}}=m_{0}2^{-{\frac {10^{6}}{1550}}}=m_{0}2^{-{\frac {t}{T}}}=m_{0}2^{-p}.}
Atomo skersmuo yra apie
10
−
10
{\displaystyle 10^{-10}}
10
30
{\displaystyle 10^{30}}
m
0
=
10
30
{\displaystyle m_{0}=10^{30}}
m
(
t
)
=
m
(
10
5
)
=
m
0
2
−
t
T
=
10
30
⋅
2
−
10
5
1550
=
10
30
⋅
2
−
64.516129
=
10
30
⋅
3.7906174467
⋅
10
−
20
=
37906174467
=
3.7906174467
⋅
10
10
{\displaystyle m(t)=m(10^{5})=m_{0}2^{-{\frac {t}{T}}}=10^{30}\cdot 2^{-{\frac {10^{5}}{1550}}}=10^{30}\cdot 2^{-64.516129}=10^{30}\cdot 3.7906174467\cdot 10^{-20}=37906174467=3.7906174467\cdot 10^{10}}
Rasime kiek liks radžio nuo pradinės masės
m
0
{\displaystyle m_{0}}
t =1000 metų, kai radžio skilimo pusamžis T =1550 metų. Randame:
m
(
t
)
=
m
(
10
3
)
=
m
0
2
−
t
T
=
m
0
⋅
2
−
1000
1550
=
m
0
⋅
2
−
0.64516129
=
0.63942130066
⋅
m
0
.
{\displaystyle m(t)=m(10^{3})=m_{0}2^{-{\frac {t}{T}}}=m_{0}\cdot 2^{-{\frac {1000}{1550}}}=m_{0}\cdot 2^{-0.64516129}=0.63942130066\cdot m_{0}.}
Dabar tarkime, kad mes žinome kiek liko radžio nuo pradinės masės
m
0
{\displaystyle m_{0}}
m
t
=
0.63942130066
⋅
m
0
.
{\displaystyle m_{t}=0.63942130066\cdot m_{0}.}
T . Tereikia išspręsti per logaritmus lygtį:
0.63942130066
⋅
m
0
=
m
0
2
−
t
T
,
{\displaystyle 0.63942130066\cdot m_{0}=m_{0}2^{-{\frac {t}{T}}},}
0.63942130066
=
2
−
t
T
,
{\displaystyle 0.63942130066=2^{-{\frac {t}{T}}},}
0.63942130066
=
2
−
1000
T
,
{\displaystyle 0.63942130066=2^{-{\frac {1000}{T}}},}
0.63942130066
T
=
(
2
−
1000
T
)
T
,
{\displaystyle 0.63942130066^{T}=(2^{-{\frac {1000}{T}}})^{T},}
0.63942130066
T
=
2
−
1000
,
{\displaystyle 0.63942130066^{T}=2^{-1000},}
0.63942130066
T
=
9.332636185
⋅
10
−
302
,
{\displaystyle 0.63942130066^{T}=9.332636185\cdot 10^{-302},}
log
10
0.63942130066
T
=
log
10
(
9.332636185
⋅
10
−
302
)
,
{\displaystyle \log _{10}0.63942130066^{T}=\log _{10}(9.332636185\cdot 10^{-302}),}
T
log
10
0.63942130066
=
log
10
(
9.332636185
⋅
10
−
302
)
,
{\displaystyle T\log _{10}0.63942130066=\log _{10}(9.332636185\cdot 10^{-302}),}
−
T
⋅
0.19421290043
=
−
301.02999566398
,
{\displaystyle -T\cdot 0.19421290043=-301.02999566398,}
T
=
301.02999566398
0.19421290043
=
1549.999999987
{\displaystyle T={\frac {301.02999566398}{0.19421290043}}=1549.999999987}
Žinoma, vietoje dešimtainio logaritmo (
log
10
x
{\displaystyle \log _{10}x}
ln
x
{\displaystyle \ln x}
Raketos uždavinys. Raketa su pradine mase
M
0
{\displaystyle M_{0}}
u
0
{\displaystyle u_{0}}
v
0
{\displaystyle v_{0}}
M
d
v
d
t
=
−
u
0
d
M
d
t
.
{\displaystyle M{dv \over dt}=-u_{0}{dM \over dt}.}
Čia
d
M
d
t
=
μ
{\displaystyle {dM \over dt}=\mu }
μ
=
c
o
n
s
t
{\displaystyle \mu =const}
M - kintanti masė raketos.
d
v
=
−
u
0
d
M
M
,
{\displaystyle dv=-u_{0}{dM \over M},}
v
=
−
u
0
ln
M
+
C
.
{\displaystyle v=-u_{0}\ln M+C.}
Konstantą C randame iš sąlygos
v
=
v
0
{\displaystyle v=v_{0}}
M
=
M
0
,
{\displaystyle M=M_{0},}
t
=
0
{\displaystyle t=0}
C
=
u
0
ln
M
0
+
v
0
,
{\displaystyle C=u_{0}\ln M_{0}+v_{0},}
v
=
u
0
ln
M
0
M
+
v
0
.
{\displaystyle v=u_{0}\ln {M_{0} \over M}+v_{0}.}
Pavyzdžiui, jeigu norima akseleruoti raketą iki šviesos greičio, kai raketos masė su kuru lygi
M
0
=
10000
(
k
g
)
,
{\displaystyle M_{0}=10000\;(kg),}
m
=
M
0
2
=
5000
(
k
g
)
,
{\displaystyle m={M_{0} \over 2}=5000\;(kg),}
E
0
=
M
0
2
c
2
=
5000
(
k
g
)
⋅
9
⋅
10
16
(
m
2
/
s
2
)
=
4.5
⋅
10
20
(
J
)
.
{\displaystyle E_{0}={M_{0} \over 2}c^{2}=5000\;(kg)\cdot 9\cdot 10^{16}\;(m^{2}/s^{2})=4.5\cdot 10^{20}\;(J).}
M
0
=
10000
(
k
g
)
{\displaystyle M_{0}=10000\;(kg)}
E
k
=
M
0
u
0
2
2
=
M
0
c
2
2
=
4.5
⋅
10
20
(
J
)
,
{\displaystyle E_{k}={M_{0}u_{0}^{2} \over 2}={M_{0}c^{2} \over 2}=4.5\cdot 10^{20}\;(J),}
m
=
5000
{\displaystyle m=5000}
2.25
⋅
10
20
(
J
)
.
{\displaystyle 2.25\cdot 10^{20}\;(J).}
m
=
M
0
2
=
M
=
5000
(
k
g
)
.
{\displaystyle m={M_{0} \over 2}=M=5000\;(kg).}
2.25
⋅
10
20
(
J
)
{\displaystyle 2.25\cdot 10^{20}\;(J)}
u
0
=
c
2
{\displaystyle u_{0}=c{\sqrt {2}}}
v
=
c
2
ln
M
0
M
=
2
⋅
3
⋅
10
8
(
m
/
s
)
⋅
ln
10000
(
k
g
)
5000
(
k
g
)
=
4.242640687
⋅
10
8
(
m
/
s
)
⋅
0.69314718
=
{\displaystyle v=c{\sqrt {2}}\ln {M_{0} \over M}={\sqrt {2}}\cdot 3\cdot 10^{8}\;(m/s)\cdot \ln {10000\;(kg) \over 5000\;(kg)}=4.242640687\cdot 10^{8}\;(m/s)\cdot 0.69314718=}
=
294077443
(
m
/
s
)
≈
2.94
⋅
10
8
(
m
/
s
)
.
{\displaystyle =294077443\;(m/s)\approx 2.94\cdot 10^{8}\;(m/s).}
Todėl, jei nepaisyti reliatyvumo teorijos, raketa pasiektų greitį 0,98c, išnaudojus visą kuro energiją anihiliacijos metu (tarkim kuro stumos naudingumas 100%), kuri sudaro pusę pradinės raketos masės.
Mastant paprastai, reikia pusė objekto masės m energijos
0.5
m
c
2
{\displaystyle 0.5mc^{2}}
c . O su visa objekto masės energija galima įgreitinti kuną iki greičio v , kuris randamas, taip:
m
v
2
2
=
m
c
2
,
{\displaystyle {mv^{2} \over 2}=mc^{2},}
v
2
=
2
c
2
{\displaystyle v^{2}=2c^{2}}
v
=
c
2
=
3
⋅
10
8
(
m
/
s
)
⋅
1.414213562
=
424264068.7
(
m
/
s
)
.
{\displaystyle v=c{\sqrt {2}}=3\cdot 10^{8}\;(m/s)\cdot 1.414213562=424264068.7\;(m/s).}
Jeigu raketos masę sudaro 99% kuras tada kuro energija apytiksliai lygi
M
0
c
2
=
10000
⋅
9
⋅
10
16
=
9
⋅
10
20
(
J
)
,
{\displaystyle M_{0}c^{2}=10000\cdot 9\cdot 10^{16}=9\cdot 10^{20}\;(J),}
4.5
⋅
10
20
{\displaystyle 4.5\cdot 10^{20}}
c
2
.
{\displaystyle c{\sqrt {2}}.}
v
=
u
0
ln
M
0
M
=
c
2
ln
10000
100
=
1953808240
(
m
/
s
)
≈
2
⋅
10
9
(
m
/
s
)
.
{\displaystyle v=u_{0}\ln {M_{0} \over M}=c{\sqrt {2}}\ln {10000 \over 100}=1953808240(m/s)\approx 2\cdot 10^{9}\;(m/s).}
Matyt, ši formulė skirta tik skaičiuoti, kai kuras sudaro visą raketos masę, todėl prieš tai reikėjo taikyti formulę:
m
v
2
4
=
m
c
2
,
{\displaystyle {mv^{2} \over 4}=mc^{2},}
v
2
=
4
c
2
,
{\displaystyle v^{2}=4c^{2},}
v
=
2
c
=
6
⋅
10
8
(
m
/
s
)
.
{\displaystyle v=2c=6\cdot 10^{8}\;(m/s).}
Ir istačius:
v
=
2
c
ln
M
0
M
=
2
⋅
3
⋅
10
8
(
m
/
s
)
⋅
ln
10000
(
k
g
)
5000
(
k
g
)
=
415888308.3
(
m
/
s
)
.
{\displaystyle v=2c\ln {M_{0} \over M}=2\cdot 3\cdot 10^{8}\;(m/s)\cdot \ln {10000\;(kg) \over 5000\;(kg)}=415888308.3\;(m/s).}
Labai artima maksimaliai įmanomai reikšmei 424264068.7 m/s.
![{\displaystyle [u(x)+v(x)]'=u'(x)+v'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28ddfee33906452d75c8e3f184f27795ee90c6d3)
![{\displaystyle [u(x)-v(x)]'=u'(x)-v'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7db82031e9cf75d1bc43b3fa8344096e70cd8293)
![{\displaystyle [u(x)v(x)]'=u'(x)v(x)+u(x)v'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faff524404c03e591f2d52a13cbaa03b9e5a7a4d)
![{\displaystyle [{\frac {u(x)}{v(x)}}]'={\frac {u'(x)v(x)-u(x)v'(x)}{v^{2}(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f130da72a854d1a95bba8ea6256384ee23fff6f)
